Best Odds?

[Forefall]

Say you have 10 armies and 3 territories linked together. What presents a better defensive scenario?

1) 8 on one territory, 1 on each of the others
2) 6 on one territory, 2 on each of the others
3) evenly split, with 4, 3, and 3 on the territories


Also, does it matter in which order they are placed? Is there an advantage to placing your largest army last or first?

This is not a strategical question, simply a mathematical one.

[alster]

Say you have 10 armies and 3 territories linked together. What presents a better defensive scenario?

1) 8 on one territory, 1 on each of the others
2) 6 on one territory, 2 on each of the others
3) evenly split, with 4, 3, and 3 on the territories


Also, does it matter in which order they are placed? Is there an advantage to placing your largest army last or first?

This is not a strategical question, simply a mathematical one.

Mathematically each dice throw is an individual throw. And the main defense gain is between 1 and 2 defending armies since you can only defend with 1 or 2 dice. Basically, all you can do is putting at least 2 armies on each territory. The armies left after having done that, doesn't matter really where you place them if facing an attacker moving in against all of them.

[Forefall]

So your saying any combination in which each of the 3 territories has at least 2 units on it are statistically equivalent. And then all of those combinations are better than any combination in which one of the territories only has 1 unit on it.

Can anyone support that?

[john1099]

this has been talked about before.
Someone said that it was best to put the larger of the armies at the end, because they have less armies attacking your strongest defense.
So, I'd probably put 2's up front and 2 in the back!

[sully800]

The best odds for surviving are 2,2,6 with the 6 as the last army that will be attacked.

As others said, the difference between 1 and 2 on defense is large because you get to add another die. And you want to have you largest force be in the back, because when the attacker gets to you they already left at lost at least 2 men (sitting as singles on the conquered territories) and therefore there is a greater chance of them not being able to use all 3 dice when they get near the end.

[alster]

The best odds for surviving are 2,2,6 with the 6 as the last army that will be attacked.

As others said, the difference between 1 and 2 on defense is large because you get to add another die. And you want to have you largest force be in the back, because when the attacker gets to you they already left at lost at least 2 men (sitting as singles on the conquered territories) and therefore there is a greater chance of them not being able to use all 3 dice when they get near the end.

Nah... I don't buy that. If the attacker cannot use all three dice at the end, it means that he has lost armies on his way. Whether he loses them on country one attacking 6 armies or on country three attacking 6 armies doesn't matter.

[ClessAlvein]

Say the attacker has 20 armies, and on all but the last country, (s)he has attacks that result in one for one.

Let's simply our calculations in the final battle for the sake of demonstrating a point. Let's say one-for-ones do not happen in the final battle, the attacker stops when (s)he has 2 armies or less left to attack, and the chances of winning is 50%.

6-2-2
Results in a final battle of 12v2; the attacker would have to lose five 3v2s in a row to not be able to attack with 3 dice. The attacker would have to win one 3v2 to win. Probability of attacker losing: (4/10)^5

2-2-6
Results in a final battle of 16v6; the attacker would have to lose seven 3v2s, not necessarily in a row, to not be able to attack with 3 dice. The attacker would have to win three 3v2s to win.

This is actually a rather difficult probability question, so you'll have to take my word that it's less likely to win 16v6 than 12v2

Bah, it's too early in the morning. I didn't prove anything, did I Maybe I'll continue this post later, when I feel like doing the math.

[Forefall]

Heh, I would love to see that proff.

[ClessAlvein]

Okay, here it is. If someone out there who's actually good at probability questions would verify my reasoning, it'd be awesome

Assumptions:
* Only 2 for 0 or 0 for 2 in the final battle
* Attacker has 50% chance of winning
* Attacker stops when (s)he no longer has 3 dice to attack with
* The armies which are not last take out armies 1 for 1 (it's only the last battle we're really interested in)

The 6-2-2 setup
Results in a final battle of 12v2. The attacker would have to lose five 3v2s in a row to not be able to attack with 3 dice. Only one 3v2 is needed to win.

Probability of attacker losing: (5/10)^5, giving us...

0.03125, or 3.125%, or 96.875% chance of winning.

Now...

The 2-2-6 setup
Results in a final battle of 16v6. The attacker would have to lose seven 3v2s, not necessarily in a row, to not be able to attack with 3 dice. Three 3v2s are needed to win, not necessarily in a row.

Case 1: Attacker rolls 7 times before losing
(5/10)^7

Case 2: Attacker rolls 8 times before losing
(7)(5/10)*(5/10)^7 (for derivation reasoning behind the 7, see note 1)

Case 3: Attacker rolls 9 times before losing
(21)(5/10)^2*(5/10)^7 (for derivation reasoning behind the 21, see note 2)

That's it for cases! If the attacker rolls more than 9 times, (s)he'll have won.

Sum it all up, and we get...

0.07617, or 7.617%, or 92.383% chance of winning.

So, leave your big armies for last, if you're going purely for defense!

(note 1) For 8 rolls, there are actually only 7 spots in which the "win" can be inserted. If it was inserted last, it would not make sense as the attacker had already "lost." Thus, it is 7, not 8.

(note 2) Again, there are only 7 spots, so it's (7!)/[(7-2)!(2!)] or 7 choose 2, giving us 21, not 8 choose 2 as one may think.

[sully800]

Nah... I don't buy that. If the attacker cannot use all three dice at the end, it means that he has lost armies on his way. Whether he loses them on country one attacking 6 armies or on country three attacking 6 armies doesn't matter.

Start believing it!

http://gamesbyemail.com/Games/Gambit/BattleOdds

Say the attacker starts with 15 armies....then you put 14 in the number of attackers.

6,2,2 gives 76.19019% chance of success

2,2,6 gives 75.30564% chance of success


The difference is slight, but since the attacker must leave one army on each country he conquers, its advantageous to have your largest force in the back. (Note that we are talking simply your best chances of survival. Hiding your largest force behind your own lines puts you in the worst position to attack on the next turn, so it may be the best defense but whether it should be used depends upon the situation.

[Ham]

I usually put down 3 to 4 in every territory.

For some reason people tend to attack you more if you bunch armies anywhere in particular.

Putting 3 or 4 everywhere makes it seem like more of a porcupine defence to me.

[Forefall]

Cool, thanks for the posts guys.

[alster]

Nah... I don't buy that. If the attacker cannot use all three dice at the end, it means that he has lost armies on his way. Whether he loses them on country one attacking 6 armies or on country three attacking 6 armies doesn't matter.

Start believing it!

http://gamesbyemail.com/Games/Gambit/BattleOdds

Say the attacker starts with 15 armies....then you put 14 in the number of attackers.

6,2,2 gives 76.19019% chance of success

2,2,6 gives 75.30564% chance of success


The difference is slight, but since the attacker must leave one army on each country he conquers, its advantageous to have your largest force in the back. (Note that we are talking simply your best chances of survival. Hiding your largest force behind your own lines puts you in the worst position to attack on the next turn, so it may be the best defense but whether it should be used depends upon the situation.

No. Why say 15 attackers? Why not say 5 attackers?

[alster]

Okay, here it is. If someone out there who's actually good at probability questions would verify my reasoning, it'd be awesome

Assumptions:
* Only 2 for 0 or 0 for 2 in the final battle
* Attacker has 50% chance of winning
* Attacker stops when (s)he no longer has 3 dice to attack with
* The armies which are not last take out armies 1 for 1 (it's only the last battle we're really interested in)

The 6-2-2 setup
Results in a final battle of 12v2. The attacker would have to lose five 3v2s in a row to not be able to attack with 3 dice. Only one 3v2 is needed to win.

Probability of attacker losing: (5/10)^5, giving us...

0.03125, or 3.125%, or 96.875% chance of winning.

Now...

The 2-2-6 setup
Results in a final battle of 16v6. The attacker would have to lose seven 3v2s, not necessarily in a row, to not be able to attack with 3 dice. Three 3v2s are needed to win, not necessarily in a row.

Case 1: Attacker rolls 7 times before losing
(5/10)^7

Case 2: Attacker rolls 8 times before losing
(7)(5/10)*(5/10)^7 (for derivation reasoning behind the 7, see note 1)

Case 3: Attacker rolls 9 times before losing
(21)(5/10)^2*(5/10)^7 (for derivation reasoning behind the 21, see note 2)

That's it for cases! If the attacker rolls more than 9 times, (s)he'll have won.

Sum it all up, and we get...

0.07617, or 7.617%, or 92.383% chance of winning.

So, leave your big armies for last, if you're going purely for defense!

(note 1) For 8 rolls, there are actually only 7 spots in which the "win" can be inserted. If it was inserted last, it would not make sense as the attacker had already "lost." Thus, it is 7, not 8.

(note 2) Again, there are only 7 spots, so it's (7!)/[(7-2)!(2!)] or 7 choose 2, giving us 21, not 8 choose 2 as one may think.

This is a flawed reasoning.

You cannot assume a final battle of either 12v.2 or 16v.6. Doing that, you're only calculating one attack.

[alster]

So your saying any combination in which each of the 3 territories has at least 2 units on it are statistically equivalent. And then all of those combinations are better than any combination in which one of the territories only has 1 unit on it.

Can anyone support that?

Well. Of course it's better to have 20 defending armies than 2 defending armies. But, each attack takes place separately. And for each time you're attacked, you have the same odds of winning whether or not you have 2 or 20 armies there. The only statistical difference - looking at each throw of the dice - is between 1 and 2.

If you're defending, spreading out and putting 2 armies on each area is the best you can do. Then, if you have guys to spare, you don't get the same leverage. But of course, the more the merrier.

[ClessAlvein]

This is a flawed reasoning.

You cannot assume a final battle of either 12v.2 or 16v.6. Doing that, you're only calculating one attack.

Actually, that's the entire point of the calculation - whether it was better to have fewer attacking armies attack fewer defending armies, or have more attacking armies attack more defending armies. I also assumed a 50% win ratio for this very purpose. Don't worry about the reasoning - it's not flawed - but the fractions I used to determine the final percentage might have been, hence my request for a mathematical check.